Chapter 7: The Laplace Transform Department of Electrical - PowerPoint PPT Presentation

Laplace and Inverse Laplace Transform: Definitions and Basics Chapter 7: The Laplace Transform Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw November 18, 2013 1 / 19 DE Lecture 10 王奕翔 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Solving an initial value problem associated with a linear differential equation: 2 Plug in the initial conditions to specify the undetermined coefficients. Question : Is there a faster way? 2 / 19 DE Lecture 10 1 General solution = complimentary solution + particular solution. 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics In Chapter 4, 5, and 6, we majorly deal with linear differential equations with continuous, differentiable, or analytic coefficients. But in real applications, sometimes this is not true. For example: Square voltage input: Periodic, Discontinuous . Question : How to solve the current? How to deal with discontinuity? 3 / 19 DE Lecture 10 E ( t ) L E ( t ) R t C 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics In this lecture we introduce a powerful tool: Laplace Transform Invented by Pierre-Simon Laplace (1749 - 1827). 4 / 19 DE Lecture 10 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Overview of the Method 5 / 19 DE Lecture 10 Find unknown y ( t ) Transformed DE Apply Laplace that satisfies DE becomes an algebraic transform and initial conditions equation in Y ( s ) Apply inverse Laplace Solution y ( t ) Solve transformed − 1 of original IVP transform equation for Y ( s ) 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics 1 Laplace and Inverse Laplace Transform: Definitions and Basics 6 / 19 DE Lecture 10 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics given that the improper integral converges. 7 / 19 Definition of the Laplace Transform Note : Use capital letters to denote transforms. DE Lecture 10 Definition For a function f ( t ) defined for t ≥ 0 , its Laplace Transfrom is defined as ∫ ∞ F ( s ) := L { f ( t ) } := e − st f ( t ) dt , 0 L L L f ( t ) − → F ( s ) , g ( t ) − → G ( s ) , y ( t ) − → Y ( s ) , etc. Note : The domain of the Laplace transform F ( s ) (that is, where the improper integral converges) depends on the function f ( t ) 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Examples of Laplace Transform 8 / 19 s s DE Lecture 10 Example Evaluate L { 1 } . ∫ ∞ ∫ T e − st (1) dt = lim e − st dt L { 1 } = T →∞ 0 0 [ − e − st ] T 1 − e − sT = lim = lim . T →∞ T →∞ 0 When does the above converge? s > 0 ! Hence, the domain of L { 1 } is s > 0 , and L { 1 } = 1 s . 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics td 9 / 19 s Examples of Laplace Transform s s DE Lecture 10 Example Evaluate L { t } . ∫ ∞ ∫ T ( − e − st ) te − st dt = lim L { t } = T →∞ 0 0 [ − te − st ] T 1 − Te − sT + 1 ∫ T = lim + s e − st dt = lim s L { 1 } . T →∞ T →∞ 0 0 When does the above converge? s > 0 ! Hence, the domain of L { t } is s > 0 , and L { t } = 1 s 2 . 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Laplace Transform of t n Proof : One way is to prove it by induction. We will show another proof after discussing the Laplace transform of the derivative of a function. 10 / 19 DE Lecture 10 n ! L { t n } = n = 0 , 1 , 2 , . . . , s > 0 s n +1 , 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Proof : 11 / 19 Laplace Transform of e at DE Lecture 10 1 { e at } L = s − a , s > a ∫ ∞ ∫ T e − ( s − a ) t dt { e at } e at e − st dt = lim L = T →∞ 0 0 [ − e − ( s − a ) t 1 − e − ( s − a ) T ] T = lim = lim s − a s − a T →∞ T →∞ 0 When does the above converge? s − a > 0 ! 1 Hence, the domain of L { e at } is s > a , and L { e at } = s − a . 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics s 12 / 19 s s s s DE Lecture 10 k s Proof : Laplace Transform of sin ( kt ) and cos ( kt ) L { sin ( kt ) } = s 2 + k 2 , L { cos ( kt ) } = s 2 + k 2 , s > 0 ∫ ∞ ∫ ∞ ( − e − st ) L { sin ( kt ) } = sin ( kt ) e − st dt = sin ( kt ) d 0 0 ∫ ∞ ] ∞ [ − sin ( kt ) e − st cos ( kt ) e − st dt = + k 0 0 ] ∞ [ − sin ( kt ) e − st = + k s L { cos ( kt ) } 0 ] ∞ [ − sin ( kt ) e − st When does the above converge? s > 0 ! = ⇒ 0 = 0 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics s 13 / 19 s . s s s s DE Lecture 10 s k Proof : Laplace Transform of sin ( kt ) and cos ( kt ) L { sin ( kt ) } = s 2 + k 2 , L { cos ( kt ) } = s 2 + k 2 , s > 0 ∫ ∞ ∫ ∞ ( − e − st ) L { cos ( kt ) } = cos ( kt ) e − st dt = cos ( kt ) d 0 0 ∫ ∞ ] ∞ [ − cos ( kt ) e − st sin ( kt ) e − st dt = − k 0 0 ] ∞ [ − cos ( kt ) e − st = − k s L { sin ( kt ) } 0 ] ∞ [ − cos ( kt ) e − st 0 = 1 When does the above converge? s > 0 ! = ⇒ 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Solve the above, we get the result: 14 / 19 s k = DE Lecture 10 Proof : k s Laplace Transform of sin ( kt ) and cos ( kt ) L { sin ( kt ) } = s 2 + k 2 , L { cos ( kt ) } = s 2 + k 2 , s > 0 { L { sin ( kt ) } = k s L { cos ( kt ) } L { cos ( kt ) } = 1 s − k s L { sin ( kt ) } s 2 − k 2 L { sin ( kt ) } = k s L { cos ( kt ) } = k s 2 L { sin ( kt ) } ⇒ s 2 + k 2 L { sin ( kt ) } = k ⇒ L { sin ( kt ) } = s 2 + k 2 s 2 s 2 = L { cos ( kt ) } = s k L { sin ( kt ) } = s 2 + k 2 . 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Proof : It can be proved by the linearity of integral. 15 / 19 s k Laplace Transform is Linear . Hence Example DE Lecture 10 Theorem L L For any α, β , f ( t ) − → F ( s ) , g ( t ) − → G ( s ) , L { α f ( t ) + β g ( t ) } = α F ( s ) + β G ( s ) Evaluate L { sinh ( kt ) } and L { cosh ( kt ) } . e kt − e − kt ) e kt + e − kt ) A: sinh ( kt ) = 1 ( , cosh ( kt ) = 1 ( 2 2 → 1 ( 1 1 ) L sinh ( kt ) − s − k − = s 2 − k 2 , s > | k | 2 s + k → 1 ( 1 1 ) L cosh ( kt ) − s − k + = s 2 − k 2 , s > | k | . 2 s + k 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics e at 16 / 19 s k s Laplace Transforms of Some Basic Functions k DE Lecture 10 t n n ! L − → s > 0 s n +1 1 L − → s > a s − a L sin ( kt ) − → s > 0 s 2 + k 2 L cos ( kt ) − → s > 0 s 2 + k 2 L sinh ( kt ) − → s > | k | s 2 − k 2 L cosh ( kt ) − → s > | k | s 2 − k 2 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Existence of Laplace Transform Theorem (Sufficient Conditions for the Existence of Laplace Transform) of exponential order , Definition lim 17 / 19 DE Lecture 10 If a function f ( t ) is piecewise continuous on [0 , ∞ ) , and then L { f ( t ) } exists for s > c for some constant c . A function f ( t ) is of exponential order if ∃ c ∈ R , M > 0 , τ > 0 such that | f ( t ) | < Me ct , ∀ t > τ. Note : If f ( t ) is of exponential order, then ∃ c ∈ R such that for s > c , t →∞ f ( t ) e − st = 0 . 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics Existence of Laplace Transform 18 / 19 exists. DE Lecture 10 Theorem (Sufficient Conditions for the Existence of Laplace Transform) of exponential order , If a function f ( t ) is piecewise continuous on [0 , ∞ ) , and then L { f ( t ) } exists for s > c for some constant c . Proof : For sufficiently large T > τ , we split the following integral: ∫ T ∫ τ ∫ T f ( t ) dt = f ( t ) e − st dt + f ( t ) e − st dt . 0 0 τ � �� � � �� � I 1 I 2 We only need to prove that I 2 converges as T → ∞ : ∫ T ∫ T ∫ T | f ( t ) e − st | dt = | f ( t ) | e − st dt ≤ Me ct e − st dt , | I 2 | ≤ τ τ τ { e ct } which converges as T → ∞ for s > c since L 王奕翔

Laplace and Inverse Laplace Transform: Definitions and Basics In this lecture, we focus on functions that are of exponential order 19 / 19 DE Lecture 10 piecewise continuous on [0 , ∞ ) , and 王奕翔